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From a sample of 45 students exam scores, it was found that there was a mean score was 75 and a standard deviation of 5. Assume the distribution of exam scores is normal.

construct and interpret a 95% confidence interval for the true mean score.

Sagot :

Answer:

The 95% confidence interval for the true mean score is [tex][73.539,76.461][/tex]

Step-by-step explanation:

Note that [tex]CI=\bar x\pm z\frac{s}{\sqrt{n}}[/tex] where [tex]\bar x[/tex] is the sample mean, [tex]z[/tex] is the upper critical value for the desired confidence level, [tex]s[/tex] is the sample standard deviation, and [tex]n[/tex] is the sample size.

Therefore, [tex]CI=75\pm 1.96\frac{5}{\sqrt{45}}\approx[73.539,76.461][/tex]

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